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[ Information | Background | The CyberTeaser | Quantum News | Miscellaneous Stuff ] Solution to CyberTeaser B216The largest sum of three sides of a die is 4 + 5 + 6 = 15. So we are looking for two numbers, each less than 15, that add up to 27. The only possibilities are 14 and 13, or 14 and 12. But in fact the sum of three visible faces of a die cannot be 13, so the total number of pips on one die is 15, with 12 pips on the other die. To see that the sum of three visible faces cannot be 13, we can argue case by case. Since 4 × 3 = 12 < 13, there must be a 5 or a 6 in a sum of 13. If the largest is 5, the only possibilities are 5 + 5 + 3 or 5 + 4 + 4, but there is only one of each number on a die. CyberTeaser Archive
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